extreme 1/(x^2)+1/y+2xy

解答

极值点

\frac{1}{x ^{2}} + \frac{1}{y} + 2 x y

极小值 (2^{\frac{1}{5}}, \frac{1}{5 8})

求解步骤

找到临界点:

(2^{\frac{1}{5}}, \frac{1}{5 8})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2}{y ^{3}}

D (x, y) = \frac{12}{x ^{4} y ^{3}} - 4

检查在每个临界点处的符号

检查临界点

(2^{\frac{1}{5}}, \frac{1}{5 8}) :

极小值

极小值 (2^{\frac{1}{5}}, \frac{1}{5 8})

e x t re m e f (x, y) = x^{4} - 2 x^{2} + y^{3} - 3 y

e x t re m e f (x) = 4 x y^{2} - 2 x^{2} y - x

e x t re m e f (x, y) = x^{4} y^{4} + 2 x^{2} y^{2} - 2 x^{2} - 2 y^{2}

e x t re m e \frac{( x + 4 ) ( x - 1 )}{3 x + 2}

e x t re m e f (x) = 4 cos (3 x) + 5